of 1 54
The Characteristic Time Of The Universe And The Ancient Alien Hypothesis!
By!
Ian Beardsley!
Copyright © 2024"
of 2 54
Contents
Abstract……………………………………………………………………………..3
Part 1: Earth/Moon/Sun System And The Proton……………………………4
Important To Point Out………………………………………………………………………..…5
Introduction…………………………………………………………………………………………..6
Earth/Moon/Sun System……………………………………………………………………..…6
Proton Radius………………………………………………………………………………………..10
Part 2: The Meter A Natural Unit………………………………………………..20
Part 3: The Solar Formulation And Solutions
For Jupiter and Saturn…………………………………………………………..28
The Solar Formulation……………………………………………………………………………29
Equating The Lunar And Solar Formulations
Yield Our 1 Second Base Unit………………………………………………………………….32
Solutions For Jupiter And Saturn……………………………………………………………37
Part 4: Modeling Habitable Star Systems…………………………………………………41
Part 5: Biological Life Part of a Universal Natural
Process In The Universe………………………………………………………………………..47
Appendix 1……………………………………………………………………………51
Appendix 2: The Data For Verifying The Equations……………………….53
of 3 54
Abstract!
It is shown here that we can find a solution to the Schrödinger wave equation for the Solar
System as planets orbiting a central star and for the moon’s of these planets. The solution is of
the same wave equation used in quantum mechanics that solves the atom as electrons around
a central nucleus. The solution for the planetary orbital systems has the same characteristic
time as the characteristic time for the proton, the primary fundamental unit of the atoms, which
is a base unit of time in terms of which everything is described. That characteristic time turns
out to be 1 second, and as such we suggest that the unit of a second is the characteristic time
of the Universe, bridging the micro-cosmos and the macro-cosmos describing them in terms of
a common set of spacetime operators. Because the second is the base unit in the Western
time measuring system that ultimately originated with the beginning of civilization in
Mesopotamia with the ancient Sumerians from their base 60 sexagesimal counting system we
look at these people who first settled down from wandering and gathering to invent agriculture,
writing, mathematics, and government.!
Because the ancient Sumerians say in their writings that they were given their knowledge by
those who came from the sky, the Anunaki, and because their mathematics lead to our system
of measuring time we have today which uses the second that we find here is the characteristic
time of the Universe in terms of modern physics, we entertain the ancient aliens hypothesis,
that posits we were visited by aliens in ancient times who perhaps kickstarted the first
civilizations which were in Mesopotamia, Egypt, and the Indus Valley, and from which
civilization throughout the world today was derived.!
We take a look at the ancient Vedic (Hindu Indian) time system and show it has a profound
connection to the that of the Mesopotamian and ancient Egyptian systems.!
In the wave solution for the planets, the mass of the Earth’s moon determines the ground state
and the algebraic condition for the Moon perfectly eclipsing the Sun as seen from the Earth is
suggested to be a possible optimum condition for a planet to support life; the lunar orbit holds
the Earth at its tilt to its orbit around the Sun preventing extreme hot and extreme cold allowing
for the seasons. The Moon appears to be some kind of a natural yardstick for measuring the
size of the Sun, which is in terms of its size, and for determining the characteristic time of the
Universe, which is in terms of its mass.!
It would be great if we could find like we did for time the characteristic lengths and masses of
the universe, and it would be great if they happened to be what our metric system uses, as is
the case with time. Something did suggest itself nicely for the unit of length, the meter, to be
Natural, and this is treated in Part 2. !
We need more observations and more complete data of other star systems to see if the
solutions here apply to them. It may be the theory presented here solves other star systems
and has related solutions for stars with habitable planets. It may be that this theory would have
applications in what could be called a project genesis where we engineer life sustaining
systems. I solve the Tau Ceti system with the theory, a star system where we have been able to
detect an Earth-like planet that is in the habitable zone of a Sun-like star. Our space-time
operators that solve the solar system and the atom are shown to perhaps solve biological
chemistry as well."
of 4 54
Part 1: Earth/Moon/Sun System And The Proton"
of 5 54
Important To Point Out
You can speak of the structure of the solar system even though it changes with time. This is
important to understand when I refer to sizes of the Moon and the planets, and their orbital
distances.
The whole object of developing a theory for the way planetary systems form is that they meet the
following criterion: They predict the Titius-Bode rule for the distribution of the planets; the
distribution gives the planetary orbital periods from Newton’s Universal Law of Gravitation. The
distribution of the planets is chiefly predicted by three factors: The inward forces of gravity from
the parent star, the outward pressure gradient from the stellar production of radiation, and the
outward inertial forces as a cloud collapses into a flat disc around the central star. These forces
separate the flat disc into rings, agglomerations of material, each ring from which a different
planet forms at its central distance from the star (it has a thickness). In a theory of planetary
formation from a primordial disc, it should predict the Titius-Bode rule for the distribution of
planets today, which was the distribution of the rings from which the planets formed.
The data in this paper is in Appendix 1 at its end.
I would like to clarify something here about the Moon:
I have shown the second in the paper is the base unit for both the Earth/Moon/Sun system as a
wave solution and the proton, making them connected to things like G, c, and h and that we
have this second when things are such that there is a perfect eclipse. The theory does not depend
on the eclipse, just the Radius of the Sun to the Radius of the Moon for the wave solution which
is the orbital radius of the Earth to the orbital radius of the Moon which is the condition of a
near perfect eclipse.
We got the second from the rotation period of the Earth at the time the moon came to perfectly
eclipse the Sun.
That is as the Earth’s day gets longer and the lunar orbit grows larger, we got the second at the
time that the Earth day was what it is during the epoch when the Moon perfectly eclipses the
Sun, 24 hours.
The near perfect eclipse is a mystery in the sense that it came to happen when anatomically
modern humans arrived on the scene, even before that, perhaps around Homo Erectus and the
beginning of the Stone Age. The Earth day was 18 hours long, long before that, 1.4 billion years
ago. Homo Erectus is around two to three million years ago.
I wrote in the first paragraph at the bottom:
"...but the theory prompts one to suggest the perfect eclipse has perhaps been a message to
humans, since hunting with stone spearpoint to present day having gone to the Moon, to tell us
we are here for reason."
I didn't say it wasn't a coincidence, but that it "prompts one to suggest" It has no bearing on the
theory.
Also, the Earth has been in the habitable zone since 4 billion years ago (0.9 AU) and that
habitable zone can continue to 1.2 AU. So we can speak of the distance to the Earth over much
time. The Earth and Sun formed about 4.6 billion years ago.
of 6 54
Introduction In order to suggest that ancient aliens had a hand in developing the way we
measure time I have to show the second, which has ancient origins, is a natural constant that
solves everything from atoms to the solar system in modern physics.!
The Natural constants, such as the constant of gravitation G, the Planck constant for the
atoms, h, and the speed of light c describe the properties of space and time. I have found they
are very conducive to our base unit for measuring time, the second. However, the base unit of
a second was given to us by the ancient Sumerians of Mesopotamia who were among the first
to settle down from wandering and gathering, from hunting with stone spearpoints to create
agriculture, writing, and mathematics. They divided the day from sunrise to sunset into 12 units
that we call today in the West, hours. As such there are 24 hours in a day from sunrise to
sunrise. They had a base 60 counting system because 60 is evenly divisible by so much, like
1,2,3,4,5,6,…the first six integers. It is the smallest number to do this. The Babylonians
adopted this base 60 (called sexagesimal) from the Sumerians, and divide their hour into 60
minutes, and the minutes into 60 seconds. We got our system from the Babylonians. Since one
day (24 hours) is the time for the Earth to make one revolution on its axis, the second comes
from dividing the Earths rotation by these numbers. But I didn’t just find that the natural
constants described the second, but that they formed spacetime operators that solve the atom
today in modern physics, and the solar system with the Schrödinger wave equation that
describes the atom in quantum mechanics. I further found that this is done with the Earth’s
moon, which seems to be a natural yardstick for measuring size and mass. This says a great
deal about the mystery known to astronomy for a long time that the Moon perfectly eclipses
the Sun as seen from the Earth. In fact, it becomes part of the solution to planetary systems,
like our solar system, for them to have life optimally possible. We know the Moon orbiting the
Earth optimizes the condition for life on Earth by holding the Earth at its tilt to its orbit around
the Sun allowing for the seasons and thus preventing extreme cold and extreme hot. Not only
does the lunar eclipse like this play a key role in the equations, but the theory prompts one to
suggest the perfect eclipse has perhaps been a message to humans, since hunting with stone
spearpoint to present day having gone to the Moon, to tell us we are here for reason.!
Earth/Moon/Sun System This second I found is the base unit of the orbital dynamics of the
solar system. I found I could create three spacetime operators, one that acts on the radius of
the proton to its mass, and the other that acts on the radius of the Sun to its mass, and one
that acts on angular momentum to speed. is Planck’s constant, is the universal constant of
gravitation, is the speed of light, is the fine structure constant, is the radius of a proton,
is the mass of a proton, is the mass of the Moon, is its radius, and the EarthDay is
the rotation period of the Earth:!
1)
2)
3)
h
G
c
α
r
p
m
p
M
m
R
m
(
1
6α
2
4πh
Gc
)
⋅
r
p
m
p
= 1secon d
(
M
m
R
m
(Ear th Da y)
)
⋅
R
⊙
M
⊙
= 1secon d
2
3
⋅
π r
p
α
4
G m
3
p
1
3
⋅
h
p
c
= 1secon d
of 7 54
They seem to suggest that the Earth orbit might by quantized in terms of the Moon and a base
unit of one second, which we will see can be considered a characteristic time of the Universe.
Equation 2 is actually 1.3 seconds but rounds to one second. However it is derived from another
equation that gives 1.2 seconds. Namely
It says one second is the Earth Day (completion of one rotation of the Earth) adjusted by the
kinetic energy of the Moon, to the kinetic energy of the Earth. As we shall see, the quantization
in terms of the Moon is exactly 1 second to nearly two places after the decimal from the ground
state of our solution to the Earth/Moon/Sun system.
We say there is a Planck constant for the solar system. We do it with a base unit of one second
because equations 1, 2, and 3 suggest we should . We suggest it is such that it is given by the
standard Planck constant for the atom, , times some constant, , and the kinetic energy of the
Earth.
4)
5)
Where
6).
Where equation 6 comes from equation 3.
We derive the value of our solar Planck constant
=
=
=
1secon d =
K E
m
K E
e
(Ear th Da y)
h
C
ℏ
⊙
= (hC )K E
e
hC = 1secon d
C =
1
3
⋅
1
α
2
c
2
3
⋅
π r
p
G m
3
p
C =
1
3
⋅
1
α
2
c
1
3
⋅
2π r
p
G m
3
p
1
3
⋅
18769
299792458
1
3
⋅
2π (0.833E − 15)
(6.67408E − 11)(1.67262E − 27)
3
1.55976565E 33
s
m
m
kg
3
⋅
s
2
kg
m
3
=
s
m
s
2
kg
2
m
2
=
s
m
⋅
s
kg ⋅ m
=
1
kg
⋅
s
2
m
2
1
C
= kg
m
2
s
2
=
1
2
mv
2
= energ y
hC = (6.62607E − 34)(1.55976565E33) = 1.03351secon ds ≈ 1.0secon d s
of 8 54
=
Now we show that our Planck constant for the solar system gives the base unit of one second for
the quantization. We call our solar Planck constant and find the wavelength for the Moon
which is the ground state for the solar system:
7)
The wavelength associated with the Moon divided by the speed of light should be 1 second if our
planetary system is quantized in terms of the Moon and a base unit of one second. We have
8)
And we see it is, so we have
9)
The solution for the orbit of the Earth around Sun with the Schrödinger wave equation can be
inferred from the solution for an electron around a proton in the a hydrogen atom with the
Schrödinger wave equation. The Schrödinger wave equation is, in spherical coordinates
10)
Its solution for the atom is as guessed by Niels Bohr before the wave equation existed:
11)
12)
hC =
(
kg
m
s
2
⋅ m ⋅ s
)
(
1
kg
⋅
s
2
m
2
)
(
kg
m
2
s
)(
1
kg
⋅
s
2
m
2
)
= secon d s
K E
earth
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E 33J
ℏ
⊙
= (hC )K E
earth
= (1.03351s)(2.7396E 33J ) = 2.8314E 33J ⋅ s
ℏ
⊙
λ
moon
=
ℏ
2
⊙
GM
3
m
=
(2.8314E 33)
2
(6.67408E − 11)(7.34763E 22kg)
3
= 3.0281E8m
λ
moon
c
=
3.0281E8m
299,792,458m /s
= 1.010secon d s
λ
moon
c
= 1secon d
−
ℏ
2
2m
[
1
r
2
∂
∂r
(
r
2
∂
∂r
)
+
1
r
2
sinθ
∂
∂θ
(
sinθ
∂
∂θ
)
+
1
r
2
sin
2
θ
∂
2
∂ϕ
2
]
ψ + V(r)ψ = E ψ
E = −
Z
2
(k
e
e
2
)
2
m
e
2ℏ
2
n
2
r
n
=
n
2
ℏ
2
Z k
e
e
2
m
e
of 9 54
is the energy for an electron orbiting protons and , is the orbital shell for an electron with
protons, the orbital number. I find the solution for the Earth around the Sun utilizes the
Moon around the Earth. This is different than with the atom because planets and moons are not
all the same size and mass like electrons and protons are, and they don’t jump from orbit to
orbit like electrons do. I find that for the Earth around the Sun
13)
14)
is the kinetic energy of the Earth, and is the planet’s orbit. is the radius of the Sun,
is the radius of the Moon’s orbit, is the mass of the Earth, is the mass of the Moon, is
the orbit number of the Earth which is 3 and is the Planck constant for the solar system.
Instead of having protons, we have the radius of the Sun normalized by the radius of
the Moon. It is has always been an amazing fact the the sizes of the Moon and the Sun are such
that given their orbital distances, the Moon as seen from the Earth perfectly eclipses the Sun.
This becomes part of the theory and we suggest it is a condition for sophisticated planets that
harbor life by writing it:
15)
That is the orbital radius of the planet (Earth) to the orbital radius of its moon (The Moon) is
about equal to the the radius of the star (The Sun) to the radius of its moon (The Moon). We say
the system is quantized by the Moon and the base unit of one second. It is a fact that the Moon
orbiting the Earth optimizes the conditions for life on Earth because it holds the Earth at its
inclination to its orbit around the Sun allowing for the seasons, preventing extreme hot and
extreme cold. Let us now see if our solar Planck constant works…
16)
=
=2.727E36J
The kinetic energy of the Earth is
The accuracy of our equation is:
E
Z
r
n
Z
n
K E
e
= n
R
⊙
R
m
⋅
G
2
M
2
e
M
3
m
2ℏ
2
⊙
r
n
=
2ℏ
2
⊙
GM
3
m
⋅
R
⊙
R
m
⋅
1
n
K E
e
r
n
R
⊙
r
m
M
e
M
m
n
ℏ
⊙
Z
R
⊙
/R
m
r
planet
r
moon
≈
R
star
R
moon
R
⊙
R
m
=
6.96E8m
1737400m
= 400.5986
K E
e
= (1.732)(400.5986)
(6.67408E − 11)
2
(5.972E 24kg)
2
(7.347673E 22kg)
3
2(2.8314E 33)
2
K E
e
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E 33J
2.727E 36J
2.7396E 33J
100 = 99.5 %
of 10 54
Which is very good. Thus we have solved the Earth/Moon/Sun System with our spacetime
operators by using them to find a Planck constant for the solar system.
We want to suggest that the base 60 dynamic applied to the rotational velocity of the Earth to
measure time has a natural property associated with the atom and the Earth/Moon/Sun orbital
system as a solution to the atom’s wave equation. The Earth day is given by, in seconds
This gives since
Thus we have as factors for the seconds per day the smallest primes 2 and 3, and the 4 of
rectangular coordinate systems, and the versatile, abundant, 60. We then suggest the second is
dynamic in terms of what the ancients gave us because 86400 seconds comes from
and this I suggest is connected to Nature because the Earth rotates at a speed
that gives from these ancient factors the duration of a second that I found is the base unit of the
atomic systems and our planetary system, in particular of our Earth/Moon/Sun system as a
solution of the Schrödinger wave equation that solves the same atomic system, which I will go
into right now.
Proton Radius We see the spacetime operators solve the atom by giving us the radius of a
proton. We set equation 3 equal to equation 1
3)
1).
These two yield
17).
I find this is close to the experimental value of the radius of a proton. I find I can arrive at this
radius of a proton another way, energy is given by Planck’s constant and frequency
We have
(
1d a y
24hrs
)(
1hr
60mi n
)(
1min
60sec
)
=
1
86400
d a y
sec
2 ⋅ 3 ⋅ 4 = 24
2 ⋅ 3 ⋅ 4 ⋅ 60 ⋅ 60 = 86400secon ds /d a y
2 ⋅ 3 ⋅ 4 ⋅ 60 ⋅ 60
2
3
⋅
π r
p
α
4
G m
3
p
1
3
⋅
h
c
= 1secon d
(
1
6α
2
4πh
Gc
)
⋅
r
p
m
p
= 1secon d
r
p
=
2
3
⋅
h
cm
p
E = h f
f = 1/s, h = J ⋅ s, h f = (J ⋅ s)(1/s) = J
E = J = Joules = energ y
of 11 54
We take the rest energy of the mass of a proton :
The frequency of a proton is
Since our theory gave us the factor of 2/3 for the radius of a proton we have:
The radius of a proton is then
This is very close to the value upon which the proton radius converged historically by two
independent methods which was 0.877E-15m. The result from our theory is
The 0.877fm was challenged in 2010 by a third experiment making it 4% smaller and was
0.842E-15m. We find it may be that the radius of a proton is actually
18)
Where is the golden ratio constant (0.618…). This is more along the lines of more recent
measurements. Both equations 17 and 18 for the radius of the proton can be right depending on
the dynamics of what is going on; the radius of a proton is not precisely defined, it is more of a
fuzzy cloud of subatomic particles. Thus we have solved the atom with our spacetime operators
by producing the radius of a proton. I began working on this theory when the proton radius was
0.833fm so it is what I have been using in this paper. We continue to be honing in on its
experimental value every year with more experiments. The vast gap between the historical
0.877fm and the 2010 0.842fm is known in physics as the proton puzzle.
Indeed if the dynamics of the factors the ancients gave us to create the duration of a second are
connected to the dynamics of stars systems then then the second should define the rotational
angular momentum of the Earth since we divide the rotation period of the Earth into these
factors to get the unit of second, and should be connected to our Planck constant for the solar
system which is in units of angular momentum as well, as is Planck’s constant for the atom.
m
p
E = m
p
c
2
f
p
=
m
p
c
2
h
m
p
c
2
h
⋅
r
p
c
=
2
3
≈ ϕ =
m
p
c
h
r
p
m
p
r
p
=
2
3
⋅
h
c
r
p
=
2
3
⋅
h
cm
p
r
p
=
2
3
⋅
6.62607E − 34
(299,792,458)(1.67262E − 27)
= 0.88094E − 15m
r
p
= ϕ ⋅
h
cm
p
= 0.816632E − 15m
ϕ
of 12 54
The angular momentum of the Earth with respect to the Sun is 2.66E40 kg m2/s. The rotational
angular momentum is 7.05E33 kg m2/s. In orbit angular momentum is given by
For a uniform rotational sphere it is given by
We found our solar system Planck constant was
This gives
19)
We are now equipped to show that the ancient Sumerians were right in dividing the rotation
period of the Earth (the day) into 24 units (the hour) because
That is
20).
Which is to say the angular momentum of the Earth to its Planck constant gives the base 60
counting in terms of the 24 hour day, the 60 of 60 minutes in an hour, and 60 seconds in a
minute, that determine our base unit of duration we call a second that happens to be, as I have
shown, the base unit of the wave solution to the atom and the Earth/Moon/Sun system.
Our equation in this paper for the Earth energy as a solution of the wave equation (eq. 13)
13).
does not depend on the Moon’s distance from the Earth, only its mass. The Moon slows the
Earth rotation and this in turn expands the Moon’s orbit, so it is getting larger, the Earth loses
energy to the Moon. The Earth day gets longer by 0.0067 hours per million years, and the
Moon’s orbit gets 3.78 cm larger per year. Equation 20
20)
only specifies divide the day into 24 units, and hours into 60 minutes and minutes into 60
seconds, regardless of what the Earth rotational velocity is. But it was more or less the same as it
is now when the Sumerians started civilization. But it may be that it holds for when the Earth
L = 2π M f r
2
L =
4
5
π M f r
2
ℏ
⊙
= 2.8314E 33J ⋅ s
L
earth
ℏ
⊙
=
7.05E 33
2.8314E 33
= 2.4899 ≈ 2.5 = 2
1
2
2.5(24) = 60
L
earth
ℏ
⊙
24 = 60
K E
e
= n
R
⊙
R
m
⋅
G
2
M
2
e
M
3
m
2ℏ
2
⊙
L
earth
ℏ
⊙
24 = 60
of 13 54
day is such that when the Moon perfectly eclipses the Sun, which we said might be a condition
for the optimization of life preventing extreme hot and cold. That is when 15 holds
15)
Which holds for today and held for the ancient Sumerians and holds for when the Earth rotation
gives the duration of a second we have today.
We want the basis set of equations for the solar system. We have
We have from equation
We have equations
We have equations
From these it becomes clear that
21)
22)
r
planet
r
moon
≈
R
star
R
moon
λ
moon
=
ℏ
2
⊙
GM
3
m
= 3.0281E8m
λ
moon
c
=
ℏ
2
⊙
GM
3
m
⋅
1
c
= 1.0secon d s
ℏ
⊙
= (hC )K E
e
hC = 1secon d
2
3
⋅
π r
p
α
4
G m
3
p
1
3
⋅
h
c
= 1secon d
(
1
6α
2
4πh
Gc
)
⋅
r
p
m
p
= 1secon d
1secon d =
K E
m
K E
e
(Ear th Da y)
ℏ
⊙
=
GM
3
m
c
K E
e
K E
e
=
GM
3
m
c
ℏ
⊙
of 14 54
23)
Thus combining equation 20 with the following…
21).
20).
Which gives
24)
=
This is very accurate to give us a second. But notice 6.262 is approximately . We see that
25)
is the circumference of a unit circle, it could be that the circumference of the Earth orbit, can
be taken as 1 (unity) and essentially we have the mystery of base 60 and the 24 hour day of the
ancient Sumerians is solved, it connects unit circle to 1 (unity).
26).
27)
The Hindu Vedic System has a day is 60 ghatika of 24 minutes each, each ghatika is divided
into 60 palas of 24 seconds each, and each pala is divided into 60 vispalas, each vispala of 0.4
ℏ
⊙
=
(
1
6α
2
⋅
r
p
m
p
4πh
Gc
)
K E
e
λ
moon
c
= 1secon d
hC = 1secon d
ℏ
⊙
= (hC )K E
e
K E
e
=
GM
3
m
c
ℏ
⊙
L
earth
ℏ
⊙
24 = 60
1secon d =
(
24
60
)
2
L
2
earth
GM
3
m
c
(
24
60
)
2
(7.05E 33)
2
(6.67408E − 11)((7.347673E 22kg)
3
(299792458m /s)
=
(
24
60
)
2
6.262sec = 1.002secon ds = 1.00secon ds
2π
1 =
(
24
60
)
2
2π
2π
2
2
=
24
60
π
cos(45
∘
) = cos(π /4) =
24
60
π
of 15 54
seconds each. So where our system has a base unit of 1 second, theirs has a base unit of 0.4
seconds, so that could be an advantage to their system, a smaller unit of time is more refined.
Further their day is divided into 60 units, ours into only 24, so their hour, the ghatika is only 24
minutes long, and ours is 60 minutes long, The proponents of this system in India say since
when working and doing chores we do a few chores in an hour, they do about one per ghatika
is 24 minutes, which makes the measure of time more manageable. That could be an
advantage I think. They say our hour is so long because the lines had to be far apart on the
Egyptian Sun Dial so the shadow cast by the Sun didn’t cross-over onto another line. But
today, with modern technology we can make clock lines marking hours closer together and
measure them with a pointer hand pointing to them without any problems, and the result is we
would have a smaller more refined hour (60 of them in a day as opposed to 24). They suggest
this method of measuring time would work better in science and engineering as well, that we
have to get away from the way sun dials had to be made in Egypt in ancient times.!
But they further point out that their system describes Nature. They say theirs are 108,000
vispalas in a day, and 108,000 vispalas in a night giving 216,000 vispalas in a 24 hour day. The
diameter of the Sun is 108 that of the Earth, and the average distance from the Sun to the
Earth is 108 solar diameters, and the average distance from the Moon to Earth is 108 lunar
diameters. 108(10)(10)(10)=108,000. Ourselves and them use base 10 counting, and that is
probably because we have ten fingers to count on.!
Conclusion
In conclusion we have that the unit of a second as derived from the Mesopotamians is a Natural
constant. As well we see the Hindu Vedic system describes Nature as well. It says
( )
( )
( )
( )
To see how this is connected to nature we would have to go into the significance of base 10
counting as we did with base 60 counting. But interesting here is that
2R
⊙
= 2(6.957E8m) = 1.3914E 9m eters
R
⊙
= Sol arRa diu s
r
e
= 1.496E11m
r
e
= Ear thOrbit
r
e
2R
⊙
= 107.52 ≈ 108
2R
moon
= 2(1.74E6m) = 3.48E6m
R
moon
= Lu n arRa diu s
r
moon
= 3.845E8m
r
moon
= Lu n ar Orbit
r
m
2R
moon
= 110 ≈ 108
(86,000sec /d a y)
2
=
43200
0.4sec /vispala
= 108,000vi spal a s /d ay
108(10)(10)(10) = 108,000
r
e
2R
⊙
= 107.52 ≈ 108
of 16 54
because
13.
where
15.
16.
Which is to say radius of the Sun to the radius of the Moon in the planetary equation plays the
role of the number of protons (Z) in an element for the energy of the electron orbits in the
atomic equation and that the Moon perfectly eclipses the Sun as see from the earth (eq 15) might
be a condition for sophisticated habitable star systems, we know the Moon allows for the
seasons making life very successful on Earth by preventing temperature extremes. So in the
Hindu Vedic system the diameters of the Sun and Moon fitting into Earth orbit and Lunar orbit
the same amount might play a similar role in a theory. You would have to go into base 10 like we
did with base 60. We have
West
24 hours
60 minutes
60 seconds
India
60 hours
24 minutes
24 seconds
A sort of inversion of things. The incredible thing here is that the wonderful equations,
equations 19 and 20:
19.
20.
unify the Mesopotamian system with the Hindu system because
r
m
2R
moon
= 110 ≈ 108
R
⊙
R
e
=
6.957E8m
6,371,000m
= 109 ≈ 108
K E
e
= n
R
⊙
R
m
⋅
G
2
M
2
e
M
3
m
2ℏ
2
⊙
r
planet
r
moon
≈
R
star
R
moon
R
⊙
R
m
=
6.96E8m
1737400m
= 400.5986
L
earth
ℏ
⊙
=
7.05E 33
2.8314E 33
= 2.4899 ≈ 2.5 = 2
1
2
L
earth
ℏ
⊙
24 = 60
of 17 54
This suggests that the Hindu system is just as Natural as the Western system.
Equation 25 that says
25)
1 hour=(60min/hour)(60sec/min)=3600sec=1 Western Hour
1 vispala=0.4sec
60 vispala=(60)(0.4sec)=24 sec=1pala
60 pala = (60)(24 sec)=1440 sec=24 min = 1 ghaticka = 1 HinduHour
28).
where is the circumference of a unit circle, and we have unity on the left, so it could be if you
combine the Indian system of measuring time with the Western system you would have a system
of units for the solar system where the equations work out nicely.
We have said the ancient Sumerians gave us the duration of a second we have today by
dividing up the rotation period of the Earth into 24 units we call hours, which the Babylonians
divided into 60 minutes and 60 seconds that they got from the Sumerian base 60. We have
found in my theory that the 1 second is a natural constant because it is the base unit of our
solution to both the solar system and the atom. We further found the Hindu’s used similar
numbers in an inverted way to the way of the West that came from the Sumerians, and that
their system relates not just to the West’s but to our theory in a dynamic way concerning the
angular momentum of the Earth and our Planck constant for the Solar system. !
The Sumerians of Mesopotamia and the Hindus of the Indus Valley Civilization were of the first
civilizations, but so was Egypt. If the second was in Mesopotamia and in India, it was only
1vi spal a = 0.4secon d s
(0.4s)(2.5) = 1secon d
1 =
(
24
60
)
2
2π
1 =
(1440sec)
2
(3600sec)
2
2π
1 =
(1gh atick a)
2
(1h our)
2
2π
1 =
Hin du Hour
2
Wester n Hour
2
2π
1 =
(60pal a)
2
(60mi n)
2
2π =
(24min)
2
(60mi n)
2
2π
1 =
((60)(24sec))
2
((60)(60sec))
2
2π =
(60pal a)
2
(60sec)
2
2π =
(24sec)
2
(60sec)
2
2π
2π
of 18 54
necessary to find it in Egypt. Since the Earth spin has been the natural clock for which the
humans first measured time since ancient times, the rising and setting of the Sun due to it, I
looked at how much distance through which the earth rotates at its equator in one second. It
is, since this distance is!
!
where theta is in radians, and the the radius of the Earth is 6.378E6m. The earth rotates
through 360 degrees per 24 hours, or per 86,400 seconds = 0.004167 degrees per second =
7.27E-5 radians. We have!
!
If the second was to exist in ancient times in Egypt we should see it in the Great Pyramids. The
three largest are in a line and are separated by!
Khufu to Khafre (0.25 miles)!
Khafre to Menkaure (0.3 miles)!
We see the second and third pyramids built, Khafre and Menkaure, which were built in the 25th
century BC and were of the the 4th dynasty of the old kingdom, are 0.3 miles apart, the
distance through which the Earth rotates in one second, the one second we found to be a
natural constant at the basis of the the atom and solar system.!
The ancient Sumerians say they got there mathematics in their writings from the Gods who
came to Earth from the sky they called the Anunaki. Some have suggested the Anunaki were
ancient aliens. The Ancient Egyptians built enormous pyramids from heavy stones weighing
tons, somehow lifted 50 feet to be stacked in near perfect mathematical proportions. This has
been a deep mystery in Archaeology. Again some suggest ancient aliens here. As well India has
its mandalas, visual painted patterns that are said to represent the sounds of chants to use to
take them into deep meditational states. The patterns found in their mandalas match the
patterns created by sound putting sand grains on a plate with transducers to vibrate them with
tones in modern acoustics studies. Again, some have suggested ancient aliens here. I have
shown all three of these civilizations have a system of measurement (the Egyptians had a 24
hour day as well) that is intrinsic to the laws of nature describing the atom and the solar system
in modern times. It can be suggested that the second was given to them by ancient aliens.!
We have focused on the aspect of the solar system that is the Earth/Moon/Sun system in this
paper. For a more complete treatment of everything here read my paper Particles Physics, The
Solar System, And Biological Systems As One System.
A really good find is the following…!
s = r θ
s = (6.378E6m)(7.27E − 5ra d ) = 464m = 0.464k m = 0.2883mi ≈ 0.3mi
of 19 54
My theory suggests the Sumerians could have ultimately got the unit of a second for
measuring time from ancient Aliens, they called the Anunaki who the say came from the sky.
The same second I have found is characteristic of our solar system and the proton. Scholars
have puzzled over the the depiction in Sumerian art of the strange watch-like bracelets around
the wrists of the Anunaki gods, who they say came from the sky and gave them mathematics.
They have twelve divisions like our clocks today have because the 12 and 6 o’clock positions
have two pointers running together as one. Click here to see the depictions:!
!
of 20 54
Part 2: The Meter A Natural Unit!
of 21 54
With the second as the characteristic time of the Universe, one might by tempted to look for
the corresponding length and mass units in the International System of Units commonly known
as the metric system, as Natural as well. Indeed I find the meter presents itself nicely as a
candidate.!
The period of a pendulum depends on the acceleration of gravity at the surface of planet where
it is oscillating. On the Earth this acceleration is g = 9.807 m/s2. And, the period of the
pendulum, T, is related to the length of the cord, , or arm, to which the mass is attached that
swings at its end, as well. It has a period given by:!
!
We have said the second is the characteristic time of the Universe, and have said it is defined
by the rotation period of the Earth and the sexagesimal mathematics given to us by the ancient
Sumerians. If the second comes to us from the rotational frequency of the earth, which is our
base unit of time, is it possible that our base unit of length in the metric system, the meter,
could come to us from The Earth’s gravity? As it would turn out, since the period of a
pendulum is given by a swing left and swing right again, that if the length of the pendulum is
one meter, that its period is 2 seconds, which means a swing left and a swing right takes about
one second. In other words, if , then then for a period of 2 seconds we have a length
because g=9.807m/s2 is approximately pi squared. It was Christian Huygens, a
Dutch mathematician and astronomer who first noticed this in the 17th century. He suggested
we define the meter so that pi squared is exactly g, which would be very close to the meter we
have today. Let’s see this:!
!
Thus we have!
!
Since the gravity at the surface of the Earth is!
!
We have!
1). !
l
T = 2π
l
g
g = π
2
l = 1meter
T = 2π
l
g
= 2π
1meter
9.807m /s
2
= 2.006373secon ds
1.0secon ds = π
1meter
g
g =
GM
e
R
2
e
1secon d = π
(1meter)
G
⋅
R
2
e
M
e
of 22 54
= !
We have equation 19 in part 1 is!
19)
And, equation 20 in part 1 is
20).
Which shows the to rotational angular momentum of the Earth is convenient for the ancient
Sumerian 24 hour day, 60 minute hour, and 60 second minute because the Solar System is
quantized in terms of . So, not only is the second defined in terms of the rotational frequency
of the Earth, but in terms of the mass of the Earth, that mass that is rotating, and the radius (or
size) of the Earth mass that is rotating. Considering our equation for the meter (equation 1):
1). !
We see it speaks of the radius of the Earth and the mass of the Earth, as well. The average
rotational angular momentum of the Earth is!
2). !
!
We have from 1!
3). !
We notice since equation 19 of part 1 is!
!
And since , angular velocity, that from equation 2:!
4). !
π
1meter(6.378E6m)
2
(6.67408E − 11)(5.972E 24kg)
= 1.0036secon d = 1.0seconds
L
earth
ℏ
⊙
=
7.05E 33
2.8314E 33
= 2.4899 ≈ 2.5 = 2
1
2
L
earth
ℏ
⊙
24 = 60
ℏ
⊙
1secon d = π
(1meter)
G
⋅
R
e
M
e
L
earth
=
4
5
πM
e
f
e
R
2
e
f
e
=
1
86400s
1meter =
(1sec)
π
2
⋅ G ⋅
M
e
R
2
e
L
earth
ℏ
⊙
= 2.5
ω
e
= 2π f
e
L
earth
=
ℏ
⊙
L
earth
M
e
R
2
e
ω
e
of 23 54
Equation 4 is!
!
5). !
From equation 3!
3). !
We can make an equation 6:!
6). !
Putting equation 6 in equation 5 gives!
7). !
We have!
8). !
Where is the cross-section of the Earth, the disc of the Earth. This is an actual thing
in physics, in climate science, where the radiation of the Sun intercepts the Earth disc to
compute the average annual temperature of the Earth. It is also what we see visually when we
look at the Earth, say from the Moon, or as it would appear from Mars. This is good because a
big part of this paper is the apparent, what the ancients observed, like the Moon perfectly
eclipsing the Sun as seen from the Earth, which plays a role in our equation for the wave
equation solution of the Earth/Moon/Sun system, because the phenomenon has an algebraic
condition. We can write 8 as!
9). !
Since from equation 20 Part 1, equation 8 can be written:!
L
2
earth
= ℏ
⊙
M
e
R
2
e
ω
e
L
earth
= ℏ
⊙
M
e
R
2
e
ω
e
1meter =
(1sec)
π
2
⋅ G ⋅
M
e
R
2
e
M
e
= (1meter)
π
2
1sec
⋅
R
2
e
G
L
earth
= R
2
e
(
π
1secon d
)
ℏ
⊙
ω
e
G
(1meter)
1secon d =
(
πR
2
e
L
earth
)
ℏ
⊙
ω
e
G
(1meter)
πR
2
e
= A
e
1secon d =
(
A
e
L
earth
)
ℏ
⊙
ω
e
G
(1meter)
L
earth
= ℏ
⊙
(
60
24
)
of 24 54
10). !
= !
too check our work so far and make sure we got the right units. Let’s check if the numbers
work, and if so, how well:!
, !
, !
= !
And it works very well. Considering equation 10, since from equation 25, part 1:!
25)
We have
And 10 becomes:
11).
If we invert 10 we have it is acceleration:
12).
Again, since , and we have:
13).
(1secon d )
2
(1meter)
= π
2
R
4
e
(
24
60
)
2
(
ω
e
ℏ
⊙
G
)
= A
2
e
(
24
60
)
2
(
ω
ℏ
⊙
G
)
(m
4
)
(
1
s
)(
s
kg ⋅ m
2
)
(
s
2
kg
kg
)
=
s
2
m
ℏ
⊙
= 2.8314E33
ω
e
= 2π /86400sec = 7.2722E − 5sec
−1
A
e
= πR
2
e
= π(6.378E6E6m)
2
= 1.278E14m
2
G = 6.674E − 11
(1.278E14m
2
)
(
24
60
)
2
(7.2722E − 5)
(2.8314E33)(6.674E − 11)
= 1.00568
s
2
m
1 =
(
24
60
)
2
2π
1
2π
=
(
24
60
)
2
(1secon d )
2
(1m eter)
=
A
2
e
2π
(
ω
e
ℏ
⊙
G
)
1
m
s
2
=
1
A
2
e
(
60
24
)
2
(
ℏ
⊙
G
ω
e
)
1
2π
=
(
24
60
)
2
ω
e
= 2π f
e
1
m
s
2
=
1
A
2
e
(
ℏ
⊙
G
f
e
)
of 25 54
Since
And
We have
14).
=
So we have unit acceleration in terms of the Earth day. This is good, it hearkens back to one of
our three spacetime operators that solve everything, equation 2 in Part 1, which is
2)
Which can equivalently can be written:
But we want that the meter itself is natural, as well. We have the second is the characteristic
time of the Universe. If the meter is natural, then the kilogram might be natural because the
density of water is is defined to be that a kilogram of water occupies a cube 10cm on each side.
But, then we would have to get into base 10 and the metric system. But we can show the meter is
natural. Equation 12 is
12).
And the ground state for the solar system in Part 1 is
Thus, the meter is
15).
1
86400s
= Ear th Rotation Frequ en c y = f
e
1
f
e
= Ear th Da y
1
m
s
2
=
(
ℏ
⊙
G
A
2
e
)
(Ear th Da y)
(2.8314E 33)(6.674E − 11)
(1.278E14)
2
86,400s = 0.9963
m
s
2
≈ 1
m
s
2
(
M
m
R
m
(Ear th Da y)
)
⋅
R
⊙
M
⊙
= 1secon d
1secon d =
K E
m
K E
e
(Ear th Da y)
1
m
s
2
=
1
A
2
e
(
60
24
)
2
(
ℏ
⊙
G
ω
e
)
λ
moon
c
=
ℏ
2
⊙
GM
3
m
⋅
1
c
= 1.0secon d s
1m eter =
(
ℏ
5
⊙
GM
6
m
⋅
1
c
2
)(
1
A
2
e
ω
e
)
(
60
24
)
2
of 26 54
= =
X =
=
We have that it works very well. How could the meter connect with ancient aliens? The answer is
there are some who believe that the meter, and indeed the kilogram, were discovered as natural
by the ancient Egyptians, and that it worked its way into Europe before the metric system was
founded in France in the 1795. That the ancient Egyptians had had it encoded in their unit of
length, the royal cubit, by way of the volume of a sphere inscribed in a cube. This was presented
in the movie “Great Pyramid K 2019” by the director Fehmi Krasniqi. It would seem according
to Quentin Leplat, an independent researcher in ancient science, twelfth century Roman
churches may have used the metric system. He took measurements and did a statistical
treatment of the data. He found these monuments used what seemed to be both the royal cubit
and the meter. He found the Roman Church of Saint Nectaire had the entrance double of the
first entrance, which was 3 meters wide. And, that the entrance behind was 4 royal cubits wide
and that the length of this Roman church was exactly 40 meters. He found The Church of the
Conques had a stone that was different than all the others and that was much larger and that
this stone measured exactly 1 meter wide.
There is an interesting thing that happens with all of this. We have said the Moon is some kind
of a natural yardstick, it describes the Sun as having a radius of 400 in our energy solution of the
wave equation, and its mass determines the characteristic time of one second in the ground state
expression for the solar system. Now that we have unit acceleration, equation 14:
we see this is good to have because the gravity at the lunar surface is 1.62 m/s2. Since 1.62~1.618
which is the golden ratio, , we have
16).
=
(
ℏ
2
⊙
G
)(
ℏ
2
⊙
M
3
m
)
(
ℏ
⊙
M
3
m
)
(
1
A
2
e
ω
e
)
(
60
24
)
2
(
1
c
2
)
(
(2.8314E 33)
2
6.674E − 11
)(
(2.8314E 33)
2
(7.3475731E 22)
3
)(
(2.8414E 33)
2
7.3476731E 22)
3
)
(
1
(1.278E14)
2
(7.2722E − 5)
)(
60
24
)
2
(
1
299792458
)
(1.201188E 77)(0.02020938)(7.1375925E − 36)(8.419232E − 25)(6.25)(8.97744E16)
1.01444m eters ≈ 1.0meters
1
m
s
2
=
(
ℏ
⊙
G
A
2
e
)
(Ear th Da y)
Φ
g
moon
= Φ
(
ℏ
⊙
G
A
2
e
)
(Ear th Da y)
(2.8314E 33))(6.674E − 11)
(1.278E14)
2
86400 = 0.99963m /s
2
≈ 1m /s
2
of 27 54
Also, one of the obvious things we could have done at the beginning of this section is to take
equation 3
And, the equation for the ground state of the Solar System
and write
17).
=
1m eter =
(1sec)
π
2
⋅ G ⋅
M
e
R
2
e
λ
moon
c
=
ℏ
2
⊙
GM
3
m
⋅
1
c
= 1.0secon d s
1m eter =
ℏ
2
⊙
M
3
m
⋅
1
c
⋅
M
e
π
2
R
2
e
(2.8314E 33)
2
(7.3474731E 22)
3
⋅
1
299792458
⋅
5.972E 24
π
2
(6.378E6)
2
= 1.00277m = 1.00m eters
of 28 54
Part 3: The Solar Formulation And Solutions For Jupiter and Saturn
of 29 54
The Solar Formulation
Our solution of the wave equation for the planets gives the kinetic energy of the Earth from the
mass of the Moon orbiting the Earth, but you could formulate based on the Earth orbiting the
Sun. In our lunar formulation we had:
1.
We remember the Moon perfectly eclipses the Sun which is to say
2.
Thus equation 1 becomes
3.
The kinetic energy of the Earth is
4.
Putting this in equation 3 gives the mass of the Sun:
5.
We recognize that the orbital velocity of the Moon is
6.
So equation 5 becomes
7.
This gives the mass of the Moon is
8.
Putting this in equation 1 yields
9.
K E
e
= 3
R
⊙
R
m
⋅
G
2
M
2
e
M
3
m
2ℏ
2
⊙
R
⊙
R
m
=
r
e
r
m
K E
e
= 3
r
e
r
m
⋅
G
2
M
2
e
M
3
m
2ℏ
2
⊙
K E
e
=
1
2
⋅
GM
⊙
M
e
r
e
M
⊙
= 3r
2
e
⋅
GM
e
r
m
⋅
M
3
m
ℏ
2
⊙
v
2
m
=
GM
e
r
m
M
⊙
= 3r
2
e
⋅ v
2
m
⋅
M
3
m
ℏ
2
⊙
M
3
m
=
M
⊙
ℏ
2
⊙
3r
2
e
v
2
m
K E
e
=
R
⊙
R
m
⋅
G
2
M
2
e
M
⊙
2r
2
e
v
2
m
of 30 54
We now multiply through by and we have
10.
Thus the Planck constant for the Sun, , in this the case the star is the Sun, is angular
momentum quantized, the angular momentum we will call , the subscript for Planck. We
have
We write for the solution of the Earth/Sun system:
11.
Let us compare this to that of an atom:
12.
We notice that in equation 11
; ; ; ;
is really . We can write 11 as
13.
We say. . That is
Let us see how accurate our equation is:
M
2
e
/M
2
e
K E
e
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2r
2
e
v
2
m
M
2
e
ℏ
⊙
L
p
p
L
p
= r
e
v
m
M
e
= r
e
v
m
M
e
= (1.496E11m)(1022m /s)(5.972E 24kg) = 9.13E 38kg ⋅
m
2
s
L
2
p
= r
2
e
v
2
m
M
2
e
= 7.4483E 77J ⋅ m
2
⋅ kg = 8.3367E 77kg
2
⋅
m
4
s
2
K E
e
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2L
2
p
E = −
Z
2
n
2
⋅
k
2
e
e
4
m
e
2ℏ
2
Z
2
n
2
→
R
⊙
R
m
k
2
e
→ G
2
e
4
→ M
4
e
m
e
→ M
⊙
ℏ
2
→ L
2
p
L
p
ℏ
⊙
K E
e
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2ℏ
2
⊙
ℏ
⊙
= ℏ
⊙
/2π
ℏ
⊙
= 9.13E 38J ⋅ s
h
⊙
= 2π ℏ
⊙
= 5.7365E 39J ⋅ s
K E
e
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2ℏ
2
⊙
of 31 54
=
=
We have that the kinetic energy of the Earth is
Our equation has an accuracy of
Which is very good.
We call the same in our solar solution. But we now want our solution for the solar
formulation. is the mass of the Earth and is the mass of the Sun. We have our solution
might be!
14. !
Where is !
We have
15.
This has an accuracy of
!
Thus the solutions to the wave equation!
R
⊙
R
m
(6.67408E − 11)
2
(5.972E 24kg)
4
(1.9891E 30kg)
2(8.3367E 77kg
2
⋅
m
4
s
2
)
R
⊙
R
m
(6.759E 30J )
R
⊙
R
m
=
6.957E8m
1737400m
= 400.426
K E
e
= 2.70655E 33J
K E
earth
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E 33J
2.70655E 33J
2.7396E 33J
= 98.79 %
L
p
ℏ
⊙
r
n
M
e
M
⊙
r
n
=
ℏ
2
⊙
GM
3
e
⋅
R
m
R
⊙
h
⊙
ℏ
⊙
= 9.13E 38J ⋅ s
r
3
=
(9.13E 38)
2
(6.67408E − 11)(5.972E 24kg)
3
⋅
1
400.5986
= 1.4638E11m
1.4638E11m
1.496E11m
100 = 97.85 %
of 32 54
!
for the solar formulations are!
13)
14)
Equating The Lunar And Solar Formulations Yield Our 1 Second Base Unit
Let us equate equation 1 with equation 13:
1.
13.
This gives:
16.
We remember that
−
ℏ
2
2m
[
1
r
2
∂
∂r
(
r
2
∂
∂r
)
+
1
r
2
sinθ
∂
∂θ
(
sinθ
∂
∂θ
)
+
1
r
2
sin
2
θ
∂
2
∂ϕ
2
]
ψ + V(r)ψ = E ψ
K E
e
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2ℏ
2
⊙
r
n
=
ℏ
2
⊙
GM
3
e
⋅
R
m
R
⊙
ℏ
⊙
= 9.13E 38J ⋅ s
h
⊙
= 2π ℏ
⊙
= 5.7365E 39J ⋅ s
K E
e
= n
R
⊙
R
m
⋅
G
2
M
2
e
M
3
m
2ℏ
2
⊙
K E
e
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2ℏ
2
⊙
3
R
⊙
R
m
⋅
G
2
M
2
e
M
3
m
2ℏ
2
⊙
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2L
2
p
L
p
=
M
2
e
M
⊙
M
3
m
3
⋅ ℏ
⊙
ℏ
⊙
= (hC )K E
p
hC = 1secon d
C =
1
3
⋅
1
α
2
c
1
3
⋅
2π r
p
G m
3
p
of 33 54
This gives
17.
We have
18.
This equates the orbital velocity of the Moon with the centripetal acceleration of the Earth in
terms of one second by way of the mass of the Earth, the mass of the Sun, the mass of the Moon,
and the orbital number of the Earth. Let us compute
19.
Let us see how well equation 18 works. at aphelion is 966 m/s and .
. We have
That is an accuracy of
Equation 18 can be written:
20.
From our equation:
1
6α
2
m
p
h 4π r
2
p
Gc
= 1.004996352secon ds
r
e
v
m
M
e
=
1
6α
2
⋅
r
p
m
p
h 4π
Gc
⋅
1
2
M
e
v
2
e
⋅
M
2
e
M
⊙
M
3
m
3
2v
m
=
v
2
e
r
e
(1secon d )
M
2
e
M
⊙
M
3
m
3
M
2
e
M
⊙
M
3
m
3
=
(5.972E 24kg)
2
(1.9891E 30kg)
(7.34763E 22kg)
3
(1.732)
= 321,331.459 ≈ 321,331
v
m
v
e
= 29,800m /s
r
e
= 1AU = 1.496E11m
2(966m /s) =
(29,800m /s)
2
1.496E11m
(1sec)(321,331.459)
1,907m /s = 1,932m /s
1907
1932
= 98.7 %
1secon d = 2r
e
v
m
v
2
e
M
3
m
3
M
2
e
M
⊙
1
6α
2
m
p
h 4π r
2
p
Gc
= 1.004996352secon ds
of 34 54
We have
21.
Since , the diameter of the Earth orbit, we have
22.
And we see the Earth/Moon/Sun system determines the radius and mass of the proton and vice
versa. We basically have
23. 62)
Where is the Earth orbital number. We have
= 7.83436E4seconds
EarthDay=(24)(60)(60)=86400 seconds,
accuracy
This last equation, equation 62, we can use to find the rotation period, or the length of the day,
of an earth-like planet in the habitable zone of any star system, so it would be very useful.
We want to turn our attention to equation 20 and write it
1
6α
2
⋅
r
p
m
p
h 4π
Gc
= 2r
e
v
m
v
2
e
⋅
M
3
m
3
M
2
e
M
⊙
2r
e
= d
e
1
6α
2
⋅
r
p
m
p
h 4π
Gc
= d
e
v
m
v
2
e
⋅
M
3
m
3
M
2
e
M
⊙
1secon d = 2v
m
r
e
v
2
e
M
3
m
3
M
2
e
M
⊙
1secon d =
K E
moon
K E
earth
Ear th Da y
1secon d =
M
m
v
2
m
M
e
v
2
m
Ear th Da y
Ear th Da y =
2r
e
v
m
M
m
M
⊙
n
n = 3
Ear th Da y =
2(1.496E11m)
966m /s
7.34763E 22kg
1.9891E 30kg
1.732
7.834E4s
86,400s
100 = 90.675 %
of 35 54
24.
We see equating solar and lunar formulations for energy yield the base unit of one second.
Equating the lunar and solar solutions for orbitals instead of the for energies, which we just did,
we have
Yields
25.
Where . The accuracy of this is
Thus the energy equations gave the equation 55:
16. !
And equating the orbital equations gives
26.
These last two yield
27.
The accuracy is
1secon d = d
e
v
m
v
2
e
M
3
m
3
M
2
e
M
⊙
2ℏ
2
⊙
GM
3
m
⋅
R
⊙
R
m
⋅
1
n
=
L
2
p
GM
3
e
⋅
R
m
R
⊙
ℏ
2
⊙
M
3
e
M
3
m
=
3
2
L
2
p
R
2
m
R
2
s
3/2 = cos(π /6)
(2.8314E 33)
2
(5.972E 24)
3
(7.34763E 22)
3
= (0.866)(9.13E 38)
2
(1737400)
2
(6.96E8)
2
4.29059E 72 = 4.5005E 72
4.29059E 72
4.5005E 72
100 = 95.3 %
L
p
=
M
2
e
M
⊙
M
3
m
3
⋅ ℏ
⊙
L
2
p
=
2 3
3
⋅
M
3
e
M
3
m
R
2
⊙
R
2
m
⋅ ℏ
2
⊙
2
R
⊙
R
m
M
e
M
⊙
= 1
of 36 54
Is 98% accuracy. The important thing that comes out of this is our base unit of a second because
we see it is also a function lunar, solar, and earth masses.
24.
(400.5986)
5.972E 24kg
1.9891E 30kg
= 1
0.98 = 1
1secon d = d
e
v
m
v
2
e
M
3
m
3
M
2
e
M
⊙
of 37 54
Solutions For Jupiter And Saturn
We have the solution for the Bohr hydrogen atom is
1
is the atomic number, the number of protons orbited, but for our system it is 1, because one
body is orbited. is the orbit number, which we will deal with later. We have our solution for the
Earth/Moon/Sun system is
2
Let is the earth mass, is the lunar mass, and is our
Planck constant for the Earth/Moon/Sun system. We compute equation 2:
=
The kinetic energy of the Earth is
This is
The Moon perfectly eclipses the Sun which means as seen from the Earth the Moon appears to
be the same size as the Sun. This is because
3
Where is the Earth orbital distance, is the Moon’s orbital distance, is the Sun’s radius,
and is the Moon’s radius. This is because though the Moon is 400 times further from the Sun
than it is from the Earth, the Sun is 400 times larger than the Moon. That is
Thus
4
We find the quantum mechanical solution to the Earth/Moon/System is
E = −
Z
2
(k
e
e
2
)
2
m
e
2ℏ
2
n
2
Z
n
E =
G
2
M
2
m
3
2h
2
⊙
n
2
M = M
e
m = M
m
h
⊙
= 2.8314E 33J ⋅ s
E =
(6.67408E − 11)
2
(5.972E 24kg)
2
(7.347673E 22kg)
3
2(2.8314E 33)
2
n
2
3.93E 30Joules
n
2
K E
e
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E 33J
1/n
2
= 697
r
e
r
m
≈
R
⊙
R
m
r
e
r
m
R
⊙
R
m
6.96E8m
1737400m
= 400.5986
R
⊙
R
m
= 400
of 38 54
5 !
for Earth orbit (Third Planet).!
!
!
We see the solar Planck constant we developed works in solving the Schrödinger wave equation
for the Earth/Moon/Sun system. My belief as to why this was never done is that it had to be
realized that the solution of the Earth kinetic energy around the Sun as a quantum mechanical
state is based on the Moon around the Earth.
Now we want to find what the wave equation solutions are for Jupiter and Saturn because they
significantly carry the majority of the mass of the solar system and thus should embody most
clearly the dynamics of the wave solution to the Solar System.
I find that as we cross the asteroid belt leaving behind the terrestrial planets, which are solid,
and go to the gas giants and ice giants, the atomic number is no longer squared and the square
root of the the orbital number moves from the numerator to the denominator. I believe this is
because the solar system here should be modeled in two parts, just as it is in theories of solar
system formation because there is a force other than just gravity of the Sun at work, which is the
radiation pressure of the Sun, which is what separates it into two parts, the terrestrial planets on
this side of the asteroid belt and the giants on the other side of the asteroid belt. The effect the
radiation pressure has is to blow the lighter elements out beyond the asteroid belt when the
solar system forms, which are gases such as hydrogen and helium, while the heavier elements
are too heavy to be blown out from the inside of the asteroid belt, allowing for the formation of
the terrestrial planets Venus, Earth, and Mars. The result is that our equation has the atomic
number of the heavier metals such as calcium for the Earth, while the equation for the giants has
the atomic numbers of the gasses. We write for these planets
6
So, for Jupiter we have (And again using the maximum orbital velocity which is at perihelion):
E = n
R
⊙
R
m
⋅
G
2
M
2
m
3
2h
2
⊙
n = 3
3(400.5986)(3.93E30J ) = 2.7268585E 33J
2.7268585E33
2.7396E33
= 99.5 %
E =
Z
n
G
2
M
2
m
3
2ℏ
2
⊙
K E
j
=
1
2
(1.89813E 27kg)(13720m /s)
2
= 1.7865E 35J
E =
Z
H
5
(6.67408E − 11)
2
(1.89813E 27kg)
2
(7.347673E 22kg)
3
2(2.8314E 33)
2
E =
Z
H
5
(3.971E 35J ) = Z
H
(1.776E 35J )
of 39 54
Jupiter is mostly composed of hydrogen gas, and secondly helium gas, so it is appropriate that
.
Our equation for Jupiter is
7
Where is the atomic number of hydrogen which is 1 proton, and for the orbital
number of Jupiter, . Now we move on to Saturn…
=
The equation for Saturn is then
8
It makes sense that Saturn would use Helium in the equation because Saturn is the next planet
after Jupiter and Jupiter uses hydrogen, and helium is the next element after hydrogen. As well,
just like Jupiter, Saturn is primarily composed of hydrogen and helium gas.
Our equation in this paper for the Earth orbit does not depend on the Moon’s distance from the
Earth, only its mass. The Moon slows the Earth rotation and this in turn expands the Moon’s
orbit, so it is getting larger, the Earth loses energy to the Moon. The Earth day gets longer by
0.0067 hours per million years, and the Moon’s orbit gets 3.78 cm larger per year. It is believed
the Moon came from a chunk knocked off the Earth from a collision with a Mars sized
protoplanet. At the time that the Moon formed it was about 4 Earth radii distant 4.5 billion
years ago. After 100 million years the Moon became tidally locked making its rotation period
equal to its orbital period keeping one face always towards the Earth. After 500 million years the
Moon was orbiting at about 20 Earth radii. It is believed that during the period of heavy
bombardment in a chaotic early solar system about 4.1 to 3.8 billion years ago a large object did
a close pass pulling the Moon further changing its orbit giving it its 5 degree offset from the
Earth’s equator. Our wave equation solution may only use the Moon’s mass but the equation for
Z
H
=
1.7865E 35J
1.776E 35J
= 1.006pr oton s ≈ 1.0pr oton s = h ydr ogen(H )
Z = Z
H
E =
Z
H
5
G
2
M
2
j
M
3
m
2ℏ
2
⊙
Z
H
n = 5
n = 5
K E
S
=
1
2
(5.683E 26kg)(10140m /s)
2
= 2.92E 34J
E =
Z
6
(6.67408E − 11)
2
(5.683E 26kg)
2
(7.347673E 22)
3
2(2.8314E 33)
2
Z
2.45
(3.5588E 34J ) = Z(1.45259E 34J )
Z(1.45259E 34J ) = (2.92E 34J )
Z = 2pr oton s = Heliu m(He)
E =
Z
He
6
G
2
M
2
s
M
3
m
2ℏ
2
⊙
of 40 54
kinetic energy of the Moon to kinetic energy of the Earth times the Earth Day equal to about one
second:
Which we connect with the equation where the proton gives one second
This holds for when the Moon was at a distance from the Earth such that it appears the same
size as the Sun, which means:
Which is when the two equations above for one second connect to our wave equation solution to
the Earth
Because .
The Moon at its inclination to the Earth in its orbit makes life possible here because it holds the
Earth at its tilt to its orbit around the Sun allowing for the seasons so the Earth doesn’t get too
extremely hot or too extremely cold. We see the Moon may be there for a reason.
K E
moon
K E
earth
(Ear th Da y) = 1.08secon ds
1
6α
2
m
p
h 4π r
2
p
Gc
= 1.004996354secon ds
r
planet
r
moon
≈
R
star
R
moon
K E
e
= n
R
⊙
R
m
⋅
G
2
M
2
e
M
3
m
2ℏ
2
⊙
R
star
R
moon
=
R
⊙
R
m
of 41 54
Part 4: Modeling Habitable Star Systems
of 42 54
Modeling Habitable Star Systems
We want to solve a habitable planetary system in general and apply it to another star system. We
must include the solution of its moon as well, because the Moon of the Earth makes life
optimally possible. However, our Moon is counter-intuitive, it breaks all of the rules of the other
moons in our solar system. One is it has a low mass for its size. It has been said that it could be
hollow. Even suggested that it is a hollow-spacecraft put there to make life as we know it
possible. In our quantum theory for the solar system, we may have discovered the science of the
engineers who may have made the Moon, and we are possibly applying their science here. Our
quantum theory for the solar system also has a solar solution as compared to a lunar solution, so
it works for star systems in general.
The perfect star system to use for our purposes here would be Tau Ceti. Tau Ceti is a star
spectrally similar to the Sun and the closest of such a G-class star, at about 12 light years distant.
Its mass is 78% of the Sun’s. Its mass is
Its radius is:
And its luminosity is
It has several unconfirmed planets with one potentially in the habitable zone. It has been the
subject of science fiction literature. It is in the constellation Cetus. From Tau Ceti the Earth
would be seen to be in the northern hemisphere constellation Bootes with an apparent
magnitude of 2.6. It is spectral class G8 V where our Sun is G2 V.
For the luminosity of a main sequence star we only need to know the mass of the star to get its
luminosity:
1
This holds in solar masses and solar luminosities. The habitable zone of a star is given by
luminosity. If the star is 100 times brighter than our Sun, then the habitable zone is 10 times
further than the Earth is from the Sun by the inverse square law.
2
In astronomical units and solar luminosities. From we can get the orbital period in Earth
years:
3
We can get the radius of the Moon of the planet if we know the mass of the planet, :
M
s
= 0.783 + / − 0.012M
⊙
R
s
= 0.793 + / − 0.004R
⊙
L
s
= 0.488 + / − 0.0010L
⊙
L = M
3.5
r
p
= L
s
r
p
T
2
p
= r
3
p
M
e
of 43 54
4
We want the mass of the Earth from the mass of the Sun. We would guess we can get it from the
mass of the Sun, the size of the Sun, and the base 60 sexagesimal system upon which we have
found everything is based. I find the relationship exists for the Earth:
5
(36)(5)=180=360/2 and 360/6=60 is base 60 sexagesimal. This gives
This is an accuracy of
We can get the orbital distance of the planet’s Moon from the planet:
6
Once we have that we can get the radius of the Moon of the planet
7
Now we can get the rotation period of the planet, its length of day:
8
Because the orbital velocity of the planet’s Moon is
9
We can use the the delocalization time get the moon’s mass, which is 1/2 the planet’s year, with
10 , See Appendix 1
2
R
⊙
R
m
M
e
M
⊙
= 1
M
e
=
5
6
2
M
⊙
R
2
⊙
r
2
e
M
e
= 0.13889(1.9891E 30kg)
6.96E8m
2
1.496E11m
2
= 5.979748E 24kg
5.972E 24
5.979748E 24
100 = 99.87 %
r
planet
r
moon
≈
R
star
R
moon
2
R
⊙
R
m
M
e
M
⊙
= 1
Ear th Da y =
2r
e
v
m
M
m
M
⊙
3
v
m
=
GM
e
r
m
τ =
m
moon
(2r
moon
)
2
ℏ
s
of 44 54
And the whole system is solved. Let us see how the theoretical luminosity appears with that
measured with equation 8.1:
It is close to the measured. The equation for predicting is approximate as the actual
values can vary a little with things like metallicity of the stars. Tau Ceti has low metallicity
compared to our Sun.
Let’s find the orbital distance of the habitable planet if it exists with equation 2:
Let’s find the orbital period of the habitable planet (It’s year) with equation 3:
,
=
Let’s get the mass of the habitable planet from equation 5:
=
Which is good because the planet, if it exists, is thought to be larger than the Earth. Let’s get the
radius of its moon from equation 4:
=
Let’s see at what distance it orbits the planet with equation 6:
L
s
= M
3.5
s
= 0.783
3.5
= 0.425L
⊙
0.448L
⊙
r
p
= 0.488 = 0.69857AU = (0.6986AU )(1.496E11m /AU ) = 1.045E11m eters
T
2
p
= r
3
p
T
2
p
= (0.69867AU )
3
T
p
= 0.584years(365.25d a ys)((24hrs)(60mi n)(60sec) = 1849638secon d s
(0.584)(365.25d a ys) = 213.306d a ys
M
p
=
5
36
M
s
R
2
s
r
2
p
M
s
= (0.783)(1.9891E 30kg) = 1.5575E 30kg
R
s
= (0.793)(6.96E8m) = 5.52 E 8m
M
p
=
5
36
(1.5575E 30kg)
(5.52E8m)
2
(1.045E11m)
2
= 6.036E 24kg
6.036E 24
5.972E 24
= 1.0107Ear th Ma sses
R
m
= 2R
s
M
p
M
s
2(5.52E 8m)
(6.036E 24kg)
(1.5575E 30kg)
= 1.53656E6m =
1.53656E6m
1.7374E6m
= 0.8844Lun arRa dii
of 45 54
=
Let’s get the orbital velocity of this moon with equation 9:
So the orbital period of the moon is:
=
Compared to that of the Earth which is 27.3 days for the sidereal month and is a 29.53day lunar
month. The Moon of Tau Ceti would have a slightly longer lunar month, too. Now we can
determine the rotation period of the planet which would be its day:
We get the mass of the moon from equation 10:
=
So the rotation period of the planet (Its Day) is from equation 8:
=
=
If the moon of Tau Ceti is to perfectly eclipse its star, like our Moon does with the Sun, to let its
inhabitants know that they are there for reason, and so as to theoretically be a condition for
optimizing life, then we have the following should hold:
r
moon
= r
planet
R
moon
R
star
= (1.045E11m)
1.53656E6m
5.52E8m
= 2.91E8m
2.91E8m
3.84E8m
= 0.7578L un arOrbital Ra dii
v
m
=
GM
p
r
m
=
(6.67408E − 11)(6.0636E 24kg
2.91E8m
= 1,176.6m /s
T
m
=
2π r
m
v
m
=
2π (.91E8m)
1,176.6m /s
= 1.554E6secon d s
(1.554E6s)
min
60sec
h our
60mi n
d a y
24h ours
= 17.987d a ys
M
m
=
ℏ
s
τ
4r
2
m
=
(2.8314E 33J ⋅ s)(0.5)(18429638s)
4(2.91E 8m)
2
= 7.7E 22kg
7.7E 22kg
7.34763E 22kg
= 1.048Ear th Moon s
Pl a n et Da y =
2r
p
v
m
M
m
M
s
3
2(1.045E11m)
1,176.6m /s
7.7E 22kg
1.5575E 30kg
= 39495.61secon d s = 658.26min = 10.971h ours
10.971
24h ours
= 0.457Ear th Da ys
r
planet
r
moon
≈
R
star
R
moon
of 46 54
We have
And it does hold. For the Earth/Moon/Sun system we get
For the Tau Ceti system the 359 is close to 360, which is a convenient amount of degrees into
which divide a circle, like we did. We see the Tau Cetians might do the same, because like us
they might choose the base 60 sexagesimal system of counting, and this 360 might influence the
way the inhabitants of a planet orbiting this star would make their calendar. Just like the 365
day year here on Earth corresponds closely to the 360 degrees of a circle. Aside from having here
on Earth the degree, we have gradians, of which there are 400 in a circle. They make it very easy
for computing right angles to a given angle. They are used more in Europe and in surveying and
architecture, though it might have some interesting connection with our 400=400.
The habitable planet of Tau Ceti would orbit it star with a period of 213 days, which would be its
year. It star is about 78% the mass of our Sun, and about 79% its size. Its habitable planet would
be a little larger than the Earth, but not much. Its Moon would be about 88% the size of our
moon, but a little bit more massive, but not much more massive. It would orbit Tau Ceti once
about every 18 days meaning there would be 11.833 months in the Tau Ceti Year, or about 12
months like we have. It would orbit the planet at about three quarters the distance ours does.
The day from sunrise to sunrise, or sunset to sunset would be about 11 hours, or close to half the
length of our day. This is if Tau Ceti was engineered for life, but it doesn’t have to have a planet
so optimally designed for life as ours. Our 24 hour day would be better for when entering the
realm of doing astronomy, because longer nights mean more complete observations of the
universe surrounding us than for the Tau Cetians. It may mean for the conditions for life to be
optimal, like here on Earth, the star would have to be as massive as the Sun, to provide a longer
year and a longer day. But we see here how the science of engineering planetary systems optimal
for life, might be done.
1.045E11m
2.91E8m
=
5.52E8m
1.53656E6m
359 = 359
400 = 400
of 47 54
Part 5: Biological Life Part of a Universal Natural Process In The Universe
of 48 54
Introduction I have a theory for the star systems, the planets and their suns, wherein such
systems are solved with the Schrödinger wave equation that is used to solve atomic systems. The
result is that star systems and atomic systems, systems on the macro scales and micro scales, are
governed by the same underlying principles. It came to pass that this theory indicated an
overlap with biological systems indicating that biological life could be part of the same idea, and
that is what I hope to pull out of that paper and develop here.
The theory for the planets, which also predicted the radius of a proton, which is one of the
fundamental particles out of which matter is made, suggested that star systems which support
life are a part of a Universal natural process. One of the conditions for star systems that support
life made use of the interesting fact that has mystified science for some time now, that the Moon
is the right size and distance from the Earth that it nearly perfectly eclipses the Sun. The reason
a moon would be necessary for life to be optimally successful on a habitable planet is that we
know our moon that orbits our Earth, makes life here successful because it holds the Earth at its
inclination to its orbit around the Sun allowing for the seasons, thus preventing extreme hot and
extreme cold.
Interestingly, the solution for the planets of the Schrödinger wave equation, was quantized in
terms of the Earth’s moon, and a base unit of one second. The Moon seems to be some sort of a
natural yardstick. The strange thing is that the basis unit of time is one second and the second
was not developed with the planets and atoms in mind, so it is strange that it lands upon the
function of a natural constant in our theory. In the theory (An Abstract Theory of Reality,
Beardsley 2024) I talk a little about why that might be. The second came to us from the Ancient
Sumerians, who invented civilization with settling down from following the herds and hunting
with stone spearpoint to invent agriculture, metallurgy, writing, and mathematics. They had a
base 60 counting system, and found it convenient then to divide the Earth day (rotation period)
into 24 hours, which in the end became adopted by the world. Their base 60 counting resulted in
the Babylonians dividing the hour into 60 minutes, and that in turn into 60 seconds, and that is
how we have the duration of a second we have today. They chose base 60 (sexagesimal) because
60 is evenly divisible by so much from which they gave us the 12 hour day and 12 hour night or
24 hour day from sunrise to sunrise, sunset to sunset. 12 times 5 is 60, also 60 times 6 is 360,
they gave us the 360 degree circle as well.
In this paper I show that the same unit of a second that describes planetary systems and atomic
system in common describes hydrocarbons, the skeletons of biological life chemistry. I further
show that it predicts the atomic radii of the hydrogen and carbon atoms from which such
skeletons are made. Hydrogen and carbon are the most abundant elements in life chemistry,
carbon is the core element upon which life is made, and hydrogen is the simplest element,
element one in the periodic table of the elements, consisting of 1 proton, and is the most
abundant element in the universe by far, and plays the dominant role in all of chemistry.
of 49 54
A Theory For Biological Hydrocarbons I have found that the basis unit of one second is
not just a Natural constant for physical systems like the atom and the planets around the Sun
(See my paper An Abstract Theory of Reality, Beardsley 2024) but for the basis of biological
life, that it is in the sixfold Nature of the chemical skeletons from which life is built, the
hydrocarbons. I found
1).
2).
Where is the fine structure constant, is the radius of a proton, is the mass of a
proton, is Planck’s constant, is the constant of gravitation, is the speed of light, is the
kinetic energy of the Moon, is the kinetic energy of the Earth, and is the
rotation period of the Earth is one day.
The first can be written:
3).
4).
From which instead of saying the left sides of these equations are seconds, we say they are
proton-seconds by not letting the units of cancel with the bodies of these equations on the
left, which are in units of mass, but rather divide into them, giving us a number of protons. I say
this is the biological because these are the hydrocarbons the backbones of biological chemistry.
We see they display sixfold symmetry. I can generate integer numbers of protons from the time
values from these equations for all of the elements with a computer program. Some results are:
1secon d =
1
6α
2
⋅
r
p
m
p
4πh
Gc
1secon d =
K E
m
K E
e
(Ear th Da y)
α = 1/137
r
p
m
p
h
G
c
K E
m
K E
e
Ear th Da y
1
α
2
m
p
h 4π r
2
p
Gc
= 6pr oton ⋅ secon ds = carbon(C )
1
α
2
m
p
h 4π r
2
p
Gc
= 1pr oton ⋅ 6secon ds = hydr ogen(H )
m
p
of 50 54
A very interesting thing here is looking at the values generated by the program, the smallest
integer value 1 second produces 6 protons (carbon) and the largest integer value 6 seconds
produces one proton (hydrogen). Beyond six seconds you have fractional protons, and the rest of
the elements heavier than carbon are formed by fractional seconds. These are the hydrocarbons
the backbones of biological chemistry. And carbon is the core element of life. We see the
duration of the base unit of measuring time, 1 second, given to us by the ancients (the base 60,
sexagesimal, system of counting of the Sumerians who invented math and writing and started
civilization), is perfect for the mathematical formulation of life chemistry. Here is the code for
the program, it finds integer solutions for time values, incremented by the program at the
discretion of the user:
#include <stdio.h>
#include <math.h>
int main(int argc, const char * argv[]) {
int n;
float value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,
c=299792458,protons[100],r=0.833E-15;
do
{
printf("By what value would you like to increment?: ");
scanf("%f", &increment);
printf("How many values would you like to calculate for t in equation 1 (no more than 100?): ");
scanf("%i", &n);
}
while (n>=101);
{
for (int i=0; i<n;i++)
{
protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));
int intpart=(int)protons[i];
float decpart=protons[i]-intpart;
t=t+increment;
if (decpart<0.25)
{ printf("%.4f protons %f seconds %f decpart \n", protons[i], t-increment, decpart);
}}}}
We have that
Since this 6 seconds is also proton-seconds we have
5). is carbon (C)
6). is hydrogen (H)
1
α
2
⋅
r
p
m
p
4πh
G c
= 18769
0.833E − 15
1.67262E − 27
4π (6.62607E − 34)
(6.67408E − 11)(299,792458)
= 6.029978s ≈ 6s
1
6pr oton s
⋅
1
α
2
⋅
r
p
m
p
4πh
G c
= 1secon d
1
1pr oton
⋅
1
α
2
⋅
r
p
m
p
4πh
G c
= 6secon d s
of 51 54
Appendix 1
If we want to prove that our planetary Planck constant is correct, the delocalization time for the
Earth should be 6 months using it, the time for the Earth to travel the width of its orbit. We want
to solve the Schrödinger wave equation for a wave packet and use the most basic thing we can
which is a Gaussian distribution. We want to then substitute for Planck’s constant that is used
for quanta and atoms our Planck-type constant (h bar solar) for the Earth/Moon/Sun system
then apply it to predict the delocalization time for the Moon in its orbit with the Earth around
the Sun.
We consider a Gaussian wave-packet at t=0:
We say that is the delocalization length and decompose the wave packet with a Fourier
transform:
is the harmonics of the wave function. We use the identity that gives the integral of a
quadratic:
Solve the equation
With the initial condition
A plane wave is the solution:
Where,
The wave-packet evolves with time as
Calculate the Gaussian integral of
h
ℏ
⊙
ψ (x,0) = Ae
−
x
2
2d
2
d
ψ (x,0) = Ae
−
x
2
2d
2
=
∫
dp
2π ℏ
ϕ
p
e
i
ℏ
px
ϕ
p
∫
∞
−∞
e
−α
2
x+βx
d x =
π
α
e
β
2
4α
iℏ∂
t
ψ (x, t) =
p
2m
ψ (x, t)
ψ (x,0) =
∫
dp ⋅ e
p
2
d
2
2ℏ
2
⋅ e
−
i
ℏ
px
e
i
ℏ
( px−ϵ( p)t)
ϵ( p) =
p
2
2m
ψ (x, t) =
∫
dp ⋅ e
p
2
d
2
2ℏ
2
⋅ e
−
i
ℏ
( px−
p
2
2m
t)
dp
of 52 54
and
The solution is:
where
is the delocalization distance, which for instance could be the width of an atom. is the
delocalization time, the average time for say an electron to traverse the diameter of the atom and
even leave it, to delocalize. If we substitute for our , and say that the delocalization distance
is for the Moon, the width of the Earth orbit, we should get a half a year for the delocalization
time, the time for the Moon and Earth to traverse the diameter of their orbit around the Sun. We
have
Where is the mass of the Moon, and is the orbital radius of the Moon. We have
Now let’s compute a half a year…
(1/2)(365.25)(24)(60)(60)=15778800 seconds
So we see our delocalization time is very close to the half year over which the Earth and Moon
travel from one position to the opposite side of the Sun. The closeness is
Thus we know our is accurate, it continues to function in a theoretical framework. The thing
about this is that it means we can predict the mass of the Moon from the Earth year. In terms of
what we said earlier that the Moon allows for life by creating the seasons, holding the Earth at
its tilt to the Sun, so we don’t go through extreme heat and cold, this suggests the Moon has a
mass that follows from Earth orbit, which is the habitable zone of the Sun, the right distance for
water to exist as liquid, and thus could be as it is for a reason, which means life might be part of
a physical process throughout the Universe, that it unfolds naturally in the evolution of star
systems.
α
2
=
d
2
2ℏ
2
+
it
2mℏ
β =
i x
ℏ
ψ
2
= ex p
[
−
x
2
d
2
⋅
1
1 + t
2
/τ
2
]
τ =
m d
2
ℏ
d
τ
ℏ
ℏ
⊙
τ =
m
moon
(2r
moon
)
2
ℏ
⊙
m
moon
r
moon
τ = 4
(7.34767E 22kg)(3.844E 8m)
2
2.8314E 33J ⋅ s
= 15338227secon d s
15338227
15778800
100 = 97.2 %
ℏ
⊙
of 53 54
Appendix 2: The Data For Verifying The Equations
(Proton Mass)
(Planck Constant)
(Proton Radius)
(Gravitational Constant)
(light speed)
(Fine Structure Constant)
Earth day=(24)(60)(60)=86,400 seconds. Using the Moon’s orbital velocity at aphelion, and
Earth’s orbital velocity at perihelion we have:
m
P
: 1.67262 × 10
−27
kg
h : 6.62607 × 10
−34
J ⋅ s
r
p
: 0.833 × 10
−15
m
G: 6.67408 × 10
−11
N
m
2
kg
2
c : 299,792,458m /s
α : 1/137
q
p
= q
e
= 1.6022E − 19coulom bs
k
e
= 8.988E 9
Nm
2
C
2
K E
moon
=
1
2
(7.347673E 22kg)(966m /s)
2
= 3.428E 28J
K E
earth
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E 33J
of 54 54
The Author
